takes 2 H's. one from OH and one from alpha carbon (there's 2 H's on the alpha) those will form a bond.

Cleavage by O3 (alkenes)

Breaks in Half (literally) and add an O at the end of the bonds.

enantiopopic

stereocenter when formed

McPBA

alkenes to epoxide. top and bottom side attack.

**the groups stay in the same place (stereocenter wise, no changes in the 3D) since it is top and bottom attack, the only thing will change is the epoxide itself. that will go on top or bottom.(plain stereo)

C=C triple bond (IR)

2250

Cleavage for alkynes by O3

if the triple bond is at the end, the C-H bond turns into CO2

triple bond breaks into a double bond with O

and a bond with an OH.

Alcohol (OH) of 1' with CrO3 /KCr2O7

Keep OH and the 2H's turn into C=O.

*Cr- Creep (keep OH)

Sharpless

Alkene with OH turns into epoxide with OH.

OH needs to be on the Right Hand corner.

(+)Det is below

(-)Det is above.

Any product OH will be going away. All products OH will be going away. Below or above attacks should be in plain stereo. because OH is going away and what other product will be going towards. leaving plain to use.

H2 and Pd-C

Adding 2H's to any double or triple bonds. what for stereocenters.

KMnO4/NaOH
&

OsO4/NaH SO3

Double bond breaks open and will turn into 2 OH bond.

C=C (IR)

1650

N-H (IR)

3500-3200

Cleavage of Alkenes by KMnO4

Breaks in Half and at the end of the double bonds will be an O. so c=o

(CH3)3COO3H/ OsO4

anti OH

Csp3-H (IR)

3000-2850

Benzene (IR)

1600-1500

C==N

2250

Csp2H (IR)

3150-3000

LiALH4 /h2H

If it is not reacting with epoxide then it will only remove the alkyl halide group. cl would become h